A brief note on Factorising and Identities . Read it before a test. Try each question at the end before you open its answer.
Factorising A common factor first. 6 x 2 − 9 x = 3 x ( 2 x − 3 ) 6x^2 - 9x = 3x(2x - 3) 6 x 2 − 9 x = 3 x ( 2 x − 3 ) . A quadratic a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c . Find two numbers that multiply to a c ac a c and add up to b b b . Split b x bx b x into them and group. A cubic. Test the divisors of the constant term. When p ( a ) = 0 p(a) = 0 p ( a ) = 0 , divide by x − a x - a x − a and factorise the quadratic left over. The identities Identity ( x ± y ) 2 (x \pm y)^2 ( x ± y ) 2 x 2 ± 2 x y + y 2 x^2 \pm 2xy + y^2 x 2 ± 2 x y + y 2 x 2 − y 2 x^2 - y^2 x 2 − y 2 ( x + y ) ( x − y ) (x + y)(x - y) ( x + y ) ( x − y ) ( x ± y ) 3 (x \pm y)^3 ( x ± y ) 3 x 3 ± y 3 ± 3 x y ( x ± y ) x^3 \pm y^3 \pm 3xy(x \pm y) x 3 ± y 3 ± 3 x y ( x ± y ) x 3 ± y 3 x^3 \pm y^3 x 3 ± y 3 ( x ± y ) ( x 2 ∓ x y + y 2 ) (x \pm y)(x^2 \mp xy + y^2) ( x ± y ) ( x 2 ∓ x y + y 2 ) x + y + z = 0 x + y + z = 0 x + y + z = 0 x 3 + y 3 + z 3 = 3 x y z x^3 + y^3 + z^3 = 3xyz x 3 + y 3 + z 3 = 3 x y z
A worked example Factorise x 3 − 3 x 2 − 10 x + 24 x^3 - 3x^2 - 10x + 24 x 3 − 3 x 2 − 10 x + 24 .
Test. p ( 2 ) = 8 − 12 − 20 + 24 = 0 p(2) = 8 - 12 - 20 + 24 = 0 p ( 2 ) = 8 − 12 − 20 + 24 = 0 , so x − 2 x - 2 x − 2 is a factor. Divide. x 3 − 3 x 2 − 10 x + 24 = ( x − 2 ) ( x 2 − x − 12 ) x^3 - 3x^2 - 10x + 24 = (x - 2)(x^2 - x - 12) x 3 − 3 x 2 − 10 x + 24 = ( x − 2 ) ( x 2 − x − 12 ) . The quadratic. x 2 − x − 12 = ( x − 4 ) ( x + 3 ) x^2 - x - 12 = (x - 4)(x + 3) x 2 − x − 12 = ( x − 4 ) ( x + 3 ) . So it is ( x − 2 ) ( x − 4 ) ( x + 3 ) (x - 2)(x - 4)(x + 3) ( x − 2 ) ( x − 4 ) ( x + 3 ) .
Practice with new numbers
A program to try The program finds the zeroes of the worked example by testing.
Read it, then run it. Do the zeroes match the factors?
1. Change the polynomial to x ** 3 - 7 * x - 6, and 24 to 6 in both places. What are the zeroes?
Mistakes to avoid Forgetting the middle term. ( x + 5 ) 2 = x 2 + 10 x + 25 (x + 5)^2 = x^2 + 10x + 25 ( x + 5 ) 2 = x 2 + 10 x + 25 , not x 2 + 25 x^2 + 25 x 2 + 25 . Wrong signs in a sum of cubes. x 3 + y 3 x^3 + y^3 x 3 + y 3 has − x y -xy − x y in its second factor. Testing numbers that cannot work. Only divisors of the constant term can be whole-number zeroes. Stopping too early. After one factor of a cubic, factorise the quadratic that is left. Check yourself Factorise x 2 − 11 x + 28 x^2 - 11x + 28 x 2 − 11 x + 28 . Stuck? Show a hint Show the answerUse an identity to find 52 × 48 52 \times 48 52 × 48 . Stuck? Show a hint Show the answerFactorise 8 x 3 − 1 8x^3 - 1 8 x 3 − 1 . Stuck? Show a hint Show the answerIf x + y + z = 0 x + y + z = 0 x + y + z = 0 , what is x 3 + y 3 + z 3 − 3 x y z x^3 + y^3 + z^3 - 3xyz x 3 + y 3 + z 3 − 3 x y z ? Stuck? Show a hint Show the answer